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FINAL EXAM LINEAR ALGEBRA QUESTIONS AND ANSWERS

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FINAL EXAM LINEAR ALGEBRA QUESTIONS AND ANSWERS

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LINEAR ALGEBRA
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LINEAR ALGEBRA









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Institución
LINEAR ALGEBRA
Grado
LINEAR ALGEBRA

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Subido en
1 de diciembre de 2024
Número de páginas
6
Escrito en
2024/2025
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Examen
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FINAL EXAM LINEAR ALGEBRA
QUESTIONS AND ANSWERS
Every matrix transformation is a linear transformation. - Answer-True

If A is a 3x5 matrix and T is a transformation defined by T(x) = Ax then the domain of T
is R3. - Answer-False

If A is an m x n matrix, then the range of the transformation x -> Ax is Rm. - Answer-
False

Every linear transformation is a matrix transformation. - Answer-False

A linear transformation preserves the operations of vector addition and scalar
multiplication. - Answer-True

A transformation T is linear if and only if
T(c1v1+c2v2)=c1T(v1)+c2T(v2)
for all vectors v1 and v2 in the domain of T and all scalars c1 and c2. - Answer-true

For vectors ~u =[52]and ~v =[ 15−10], the inner product, of ~u and ~v is A. 55 B. 65 C.
75 D. 85 - Answer-55

List the following vectors from greatest to least in terms of their magnitude. A.[1 3 -2] B.
[0 4 2 1] C.[2 5] D. [1 0 -2 0 1] E.[0 0 0 ] - Answer-Think of Phys 2 take the square root
and square each number to find the magnitude. The order is C,B,A,D,E

Consider the following set of vectors,~u1 = 3−30 , ~u2 =22−1 , and ~u3 =114First, verify
that the given set is an orthogonal basis for R3. Then for ~y =5−31, find c1, c2 and c3
such that~y = c1~u1 + c2~u2 + c3~u3. - Answer-To proof if they Are orthogonal or not
do the dot product of all the vectors and if its 0 its orthogonal.

Then to find c1,c2,c3 do the formula C1=((y*u1)/(u1*u1)) for each C

c1=4/3 c2=1/3 c3=1/3

Consider the matrix A =1 0 −3 2
0 1 −5 4
3 −2 1 −2.
(a) (4 points) Find a basis for ColA
(b)Find a basis for NulA - Answer-a) To find the colA first row reduce the given matrix
and then take not of each of the cols that pivot then write just the col that pivot but from
the og matrix
b) from the row reducting in a write the answer in parametric form

, We know by theorem that for any matrix A, Row(A)⊥ = Nul(A) and Col(A)⊥ =
Nul(AT )This theorem isn't used often but can be used to help find an orthogonal
complement for a given space. Consider the space, W = Span1233 ,4630 ,2−3−30
(a) (4 points) Do the given vectors form a basis for W ? Why or why not? A. Yes B. No

(B) If the given set of vectors formed a matrix, A, then W wo - Answer-A) The answer is
A since you are able to roq reduce the matrix and that is the basis
B) the answer is B
C) first find the transpose of A then row reduce. Then write it in parametric vector form
and that is the basis for Nu l(A transpose) which is parallel to W

Consider the system of equations A~x = ~b where
A =1 1 0
1 0 −1
011
−1 1 −1
and ~b =3456

(a) (5 points) Show that ~b is not in ColA
b) (3 points) Show that columns of A are orthogonal.
(c) (5 points) We still would like to find some approximate solution to A~x = ~b. Find the
vector, ˆb, in ColA such that ˆb is the best approximation to ~b
(d) (5 points) Finally, calculate the distance between the original given ~b and ColA -
Answer-a) add martix and b together and row reduce if its inconsistant then vector b isnt
in ColA

b) seperate a into cols and take the dot product of each row if they equal 0 its
orthogonal;

c) find b _hat by projcolAb_vector by formula b = proj ColA~b = ~b · ~a1~a1 · ~a1~a1 +
~b · ~a2~a2 · ~a2~a2 + ~b · ~a3~a3 · ~a3~a3

d) take the magntuide of b_vector- b_hat

Let A =3 −5 1
111
−1 5 −2
3 −7 8.
(a) (8 points) Use the Gram-Schmidt algorithm to find an orthogonal basis for the
column space of A. Keep thefirst column of A and use the algorithm in the order of the
columns - Answer-Find V1,v2,v3 and seperate the cols of A to be a1,a2,a3
v1=a1
then find a2_hat by some weird equation I cant explain yet
get v2=a2-a2_hat
for a3_hat do the same equation
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